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Math Help - Integral problem

  1. #1
    Newbie LaraSoft's Avatar
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    Unhappy Integral problem

    Hello!
    Can someone help me with this problem (advanced section):

    Calculate the indefinite integral \int[(1+\sin(x))/(1-\sin(x))^2]dx, not using a universal substitution \tan(x/2)=t .

    Thanks in advance!
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Quote Originally Posted by LaraSoft View Post
    Hello!
    Can someone help me with this problem (advanced section):

    Calculate the indefinite integral \int[(1+\sin(x))/(1-\sin(x))^2]dx, not using a universal substitution \tan(x/2)=t .

    Thanks in advance!
    Multiply numerator and denominator by (1 + \sin x)^2 and use identities on the denominator to obtain \cos^4x expand and integrate term by term.
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  3. #3
    Senior Member DeMath's Avatar
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    Or so

    I = \int {\frac{{1 + \sin x}}{{{{\left( {1 - \sin x} \right)}^2}}}dx}  = \int {\frac{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}}{{\left( {1 - \sin x} \right){{\left( {1 - \sin x} \right)}^2}}}}  = \int {\frac{{{{\cos }^2}x}}{{{{\left( {1 - \sin x} \right)}^3}}}dx}  =

    = \frac{1}{2}\int {\cos xd\left( {{{\left( {1 - \sin x} \right)}^{ - 2}}} \right)}  = \frac{{\cos x}}{{2{{\left( {1 - \sin x} \right)}^2}}} + \frac{1}{2}\int {\frac{{\sin x}}{{{{\left( {1 - \sin x} \right)}^2}}}dx}  =

    = \frac{{\cos x}}{{2{{\left( {1 - \sin x} \right)}^2}}} + \frac{1}{4}\int {\frac{{1 + \sin x - \left( {1 - \sin x} \right)}}{{{{\left( {1 - \sin x} \right)}^2}}}dx}  =

    = \frac{{\cos x}}{{2{{\left( {1 - \sin x} \right)}^2}}} + \frac{1}<br />
{4}\underbrace {\int {\frac{{1 + \sin x}}{{{{\left( {1 - \sin x} \right)}^2}}}dx} }_I - \frac{1}{4}\int {\frac{{dx}}{{1 - \sin x}}}  \Leftrightarrow

    \Leftrightarrow I - \frac{1}{4}I = \frac{{\cos x}}{{2{{\left( {1 - \sin x} \right)}^2}}} - \frac{1}{4}\int {\frac{{dx}}{{1 - \sin x}}}  \Leftrightarrow

    \Leftrightarrow I = \frac{{2\cos x}}{{3{{\left( {1 - \sin x} \right)}^2}}} - \frac{1}{3}\int {\frac{{dx}}{{1 - \sin x}}}.

    \int {\frac{{dx}}{{1 - \sin x}}}  = \int {\frac{{1 + \sin x}}{{1 - {{\sin }^2}x}}dx}  = \int {\frac{{1 + \sin x}}{{{{\cos }^2}x}}dx}  = \int {\frac{{dx}}{{{{\cos }^2}x}}}  + \int {\frac{{\sin x}}{{{{\cos }^2}x}}dx}  =

    = \tan x - \int {\frac{{d\left( {\cos x} \right)}}{{{{\cos }^2}x}}}  = \tan x + \frac{1}{{\cos x}} + C = \tan x + \sec x + C.

    I = \frac{{2\cos x}}{{3{{\left( {1 - \sin x} \right)}^2}}} - \frac{1}{3}\left( {\tan x + \sec x} \right) + C.
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